InterviewSolution
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InterviewSolution
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Two non-negative integers x and y are chosen at random with replacement. The probability that `x^(2)+y^(2)` is divisible by 10, isA. `(9)/(50)`B. `(9)/(25)`C. `(3)/(50)`D. `(6)/(25)` |
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Answer» Correct Answer - A By division algorithm, we have `x=10x_(1)+a_(1) " and " y=10y_(1)++b_(1)`, where `x_(1),y_(1),a_(1),b_(1)` are integers such that `0 le a_(1) le 9 " and " 0 le b_(1) le 9`. `therefore x^(2)+y^(2)=(10x_(1)+a_(1))^(2)+(10y_(1)+b_(1))^(2)` `=100(x_(1)^(2)+y_(1)^(2))+20(a_(1)x_(1)+b_(1)y_(1))+(a_(1)^(2)+b_(1)^(2))` It is evident from this expression that `x^(2)+y^(2)` will be divisible by 10 iff `a_(1)^(2)+b_(1)^(2)` is divisible by 10. Now, there are 10 chocies each for `a_(1) " and" b_(1)`, so that there are `10xx10=100` ways of choosing them. Now, `a_(1)^(2)+b_(1)^(2)` will be divisible by 10 in each of the following cases : (0,0),(1,3),(1,7),(2,4),(2,6),(3,1),(3,9),(4,2),(4,8),(5,5),(6,2),(6,8),(7,1),(7,9),(8,4),(8,6),(9,3),(9,7) `therefore` Favourable number of elementary events =18. Hence, required probability `=(18)/(100)=(9)/(50)` |
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